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Section 1.3 Convergence Tests

Subsection 1.3.1 Direction Comparison

Here are some comparison examples.

\begin{equation*} \sum_{n=3}^\infty \frac{1}{n-2} \end{equation*}

The terms \(\frac{1}{n-2}\) are larger that \(\frac{1}{n}\) and the harmonic series \(\sum \frac{1}{n}\) is divergent, so this series is also divergent.

\begin{equation*} \sum_{n=1}^\infty \frac{1}{3^n + 4n + 1} \end{equation*}

The terms \(\frac{1}{3^n + 4n + 1}\) are smaller that \(\frac{1}{3^n}\text{.}\) These term \(\frac{1}{3^n}\) are the terms of a geometric series with common ratio \(\frac{1}{3}\text{,}\) which converges. Therefore, this series converges.

\begin{equation*} \sum_{n=1}^\infty \frac{n+1}{n^2} \end{equation*}

The terms \(\frac{n+1}{n^2}\) are larger than \(\frac{1}{n}\text{.}\) The latter are the terms of the divergent harmonic series, so this series diverges.

\begin{equation*} \sum_{n=1}^\infty \frac{2^n}{n!} \end{equation*}

For \(n \geq 4\) we have

\begin{equation*} \frac{2^n}{n!} = \frac{2}{1} \frac{2}{2} \frac{2}{3} \ldots \leq \frac{2}{3} \left( \frac{1}{2} \right)^{n-4} \end{equation*}

Therefore, we can compare to a geometric series with common ratio \(\frac{1}{2}\text{,}\) which converges. Therefore, this series also converges (and converges to \(e^2\text{,}\) as it happens).

As the last paranthetical comment hinted, comparison doesn't actually give the value of the series. These comparison arguments are very useful for determining convergence and divergence, but they don't calculate exact values. Also, in the last example the comparison only held for \(n \geq 4\) instead of all \(n \in \NN\text{.}\) This is typical and perfectly acceptable; for everything involving series other than calculating the exact value, we only need to consider the long term behaviour. For comparison, it is enough that \(a_n \lt b_n\) for all \(n\) past some finite fixed value.

Subsection 1.3.2 Asymptotic Comparison

In addition to the exact comparisons listed above, we can also compare asymptotically. Asymptotic comparison is particularly useful, since we don't actually have to calculate the inequalities.

In Example 1.3.2, we could have simply said that \(\frac{1}{n-2}\) is asymptotically the same order as \(\frac{1}{n}\text{,}\) and likewsie for \(\frac{n+1}{n^2}\text{.}\) Asymptotic comparison is often easier since we don't need to explicitly construct the necessary inequality.

As an example for both asymptotic comparison and conditional convergence, here are three alternating series. They are all convergent by the alternating series test. Comparison to geometric series or a \(\zeta\) series is used to check their absolute convergence.

\begin{align*} \amp \sum_{n=1}^\infty \frac{(-1)^n}{n^6} \amp \amp \text{ absolutely convergent by asymptotic comparison to } \frac{1}{n^6}\\ \amp \sum_{n=1}^\infty \frac{(-1)^n \arctan n}{n^2} \amp \amp \text{ absolutely convergent by asymptotic comparison to } \frac{1}{n^2}\\ \amp \sum_{n=2}^\infty \frac{(-1)^n}{\ln n} \amp \amp \text{ conditionally convergent by asymptotic comparison to } \frac{1}{\ln n} \end{align*}

In the last compoarision, \(\frac{1}{\ln n} > \frac{1}{n}\text{,}\) so the asymptotic order of \(\frac{(-1)^n}{\ln n}\) is that of a divergent series, growing faster than the harmonic series.

Subsection 1.3.3 The Integral Test

Note that the integral and the resulting series will sum to different numbers: this test doesn't calculate the value of the sum. It just tells us whether the sum is convergent.

As promised, the integral tests allows us to prove the that \(\zeta\) series converges if and only if \(p>1\text{.}\)

\begin{align*} \int_1^\infty \frac{1}{x} dx = \infty \amp \implies \sum_{n=1}^\infty \frac{1}{n} = \infty\\ \int_1^\infty \frac{1}{x^p} dx \lt \infty \amp \implies \sum_{n=1}^\infty \frac{1}{n^p} \lt \infty \text{ for } p > 1 \end{align*}

Subsection 1.3.4 The Ratio and Root Tests

Here are two final tests.

The ratio test is useful for powers and particularly for factorials. The root test is obviously useful for powers.

Subsection 1.3.5 Testing Strategies

There are many approaches to testing the convergence of a series: looking at partial sums, testing for divergence, comparison, asymptotic comparison, the alternating series test, the integral test, the ratio test, and the root test. It is difficult to know where to start and which tests or techniques to use. Here are some pointers and strategies to help you.

  • Looking at a series for asymptotic order is often the easiest first step. The main comparisons are with geometric series and \(\zeta\)-series.

  • Using the test for divergence is also often an easy first step. If the terms do not tend to \(0\text{,}\) the series cannot converge.

  • If the series is an alternating series, the alternating series test is likely the easiest approach.

  • The integral test is often the best approach if the series involves complicated functions, such as exponentials, logarithms or trigonometric functions.

  • The ratio test is often the best approach when the terms involve factorials. It is also very useful for terms which have the index in the exponent.

  • The root test is rarely used. It also helps when the index is in the exponent, but most of those cases can also be done with the ratio test.

A final important observation is that convergence only cares about the long-term behaviour of the series. Any finite pieces at the start are negligible. This is a nice observation for many of the tests: comparisons only need to work eventually, integrals can be taken on \([a,\infty)\) for some \(a>0\text{,}\) and a series which eventually becomes an alternating series can use the alternating series test.

For an extreme example, consider this series:

\begin{equation*} \sum_{n=1}^{10^{300}} (n^2 +n)^{75} + \sum_{n=10^{300} + 1}^\infty \frac{1}{n^2} \end{equation*}

The first \(10^{300}\) terms of this series are enormous numbers and their sum is simply ridiculous. However, the series is eventually is a \(\zeta\)-series with \(p=2\text{,}\) which converges. Therefore, this sum is finite. The ridiculous number we get from the first \(10^{300}\) terms is very, very large, but certainly finite. Any very, very large number is negligible when asking about infinity.

Subsection 1.3.6 Testing Examples

Now that we have all the tools at our disposal, here are a bunch of examples.

\begin{equation*} \sum_{n=1}^\infty n^{-\frac{2}{3}} \end{equation*}

The terms are \(\frac{1}{n^{\frac{2}{3}}}\text{,}\) so this is a \(\zeta\) series. Since \(\frac{2}{3} \lt 1\text{,}\) this diverges.

\begin{equation*} \sum_{k=1}^\infty \frac{2^k}{e^k} \end{equation*}

The terms are \(\left( \frac{2}{e} \right)^k\text{,}\) so this is a geometric series. \(\frac{2}{e} \lt 1\) so converges.

\begin{equation*} \sum_{k=1}^\infty \frac{(-1)^k}{k^2-1} \end{equation*}

This is an alternating series with decreasing terms, so it converges.

\begin{equation*} \sum_{n=1}^\infty \sin \left( \frac{n^2+1}{n} \right) \end{equation*}

The terms do not tend to zero, so the series is divergent by the test for divergence.

\begin{equation*} \sum_{k=1}^\infty \frac{1}{k^2-1} \end{equation*}

This is asymptotically \(\frac{1}{k^2}\text{,}\) which is a convergent \(\zeta\) series.

\begin{equation*} \sum_{k=1}^\infty \frac{(-1)^k k}{e^k} \end{equation*}

This is an alternating series with decreasing terms, so it converges.

\begin{equation*} \sum_{n=1}^\infty \frac{3}{2 + e^n} \end{equation*}

This is asymptotically \(\frac{1}{e^n}\text{,}\) which is a convergent geometric series.

\begin{equation*} \sum_{k=1}^\infty \frac{k\sqrt{k}}{k^3} \end{equation*}

This is asymptitically \(\frac{1}{k^{\frac{3}{2}}}\text{,}\) which is a convergent \(\zeta\) series.

\begin{equation*} \sum_{n=1}^\infty \frac{(-1)^n n}{n^3+4} \end{equation*}

This is an alternating series. The terms are decreasing, so the alternating series test gives convergence. In addition, the absolute value of the terms is \(\frac{n}{n^3+4}\) which is asymptotically \(\frac{1}{n^2}\text{.}\) That converges, so the series is absolutely convergent and can be rearranged without changing the value.

\begin{equation*} \sum_{k=1}^\infty \frac{2^k k!}{k^k} \end{equation*}

The factorial suggests that the ratio test is the best approach.

\begin{align*} \lim_{k \rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right| \amp = \lim_{k \rightarrow \infty} \frac{2^{k+1} (k+1)! k^k}{2^k k! (k+1)^{k+1}} = \lim_{k \rightarrow \infty} 2 (k+1) \left( \frac{k}{k+1} \right)^k \frac{1}{k+1}\\ \amp = \lim_{k \rightarrow \infty} 2 \left( \frac{k}{k+1} \right)^k = \frac{2}{e} \lt 1 \end{align*}

By the ratio test, this is convergent.

\begin{equation*} \sum_{k=1}^\infty k^5 e^{-k} \end{equation*}

We use the ratio test:

\begin{align*} \lim_{k \rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right| \amp = \lim_{k \rightarrow \infty} \frac{(k+1)^5 e^k}{k^5 e^{k+1}} = \lim_{k \rightarrow \infty} \frac{1}{e} \left( \frac{k+1}{k} \right)^5 \\ \amp = \lim_{k \rightarrow \infty} \frac{1}{e} \left( 1 + \frac{1}{k} \right)^5 = \frac{1}{e} \lt 1 \end{align*}

By the ratio test, this is convergent.

\begin{equation*} \sum_{n=1}^\infty \frac{\ln n^2}{n^2} \end{equation*}

We use the integral test, with \(u = ln x\text{.}\)

\begin{align*} \int_1^\infty \frac{\ln x^2}{x^2} dx \amp = 2 \int_1^\infty \frac{\ln x}{x^2} dx = 2 \int_0^\infty u e^{-u} du \\ \amp = \left. - ue^{-u} \right|_1^\infty + 2 \int_0^\infty e^{-u} = 0 - \left. e^{-u} \right|_1^\infty = 1 \leq \infty \end{align*}

By the integral test, this is convergent.

\begin{equation*} \sum_{n=1}^\infty \frac{(n!)^2}{(2n+4)!} \end{equation*}

The presence of a factorial means that ratio test is probably the best.

\begin{align*} \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| \amp = \lim_{n \rightarrow \infty} \frac{\frac{((n + 1)!)^2}{(2(n + 1) + 4)!}}{\frac{(n!)^2}{(2(n) + 4)!}} = \lim_{n \rightarrow \infty} \frac{(n+1)^2}{(2n+5)(2n+6}\\ \amp = \lim_{n \rightarrow \infty} \frac{n^2+2n+1}{4n^2+ 22n + 30} = \frac{1}{4} \lt 1 \end{align*}

The limit is less than 1, so the series is convergent.

\begin{equation*} \sum_{n=1}^\infty \frac{n!}{e^{n^2}} \end{equation*}

We have factorials again, so ratio test is likely the best choice.

\begin{align*} \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| \amp = \lim_{n \rightarrow \infty} \frac{\frac{(n+1)!}{e^{(n+1)^2}}}{\frac{n!}{e^{n^2}}} = \lim_{n \rightarrow \infty} \frac{(n+1)e^{n^2}}{e^{n^2+2n+1}}\\ \amp = \lim_{n \rightarrow \infty} \frac{(n+1)e^{n^2}}{e^{n^2} e^{2n} e} = \lim_{n \rightarrow \infty} \frac{n+1}{e^{2n+1}} = 0 \end{align*}

The limit is less than 1, so the series is convergent.

\begin{equation*} \sum_{n=1}^\infty \frac{\ln n}{\sqrt[3]{n}} \end{equation*}

The integral test if most appropriate here, even though the interal is difficult.

\begin{align*} \int_1^\infty \frac{\ln x}{\sqrt[3]{x}} \amp = \int_1^\infty \ln x x^{\frac{-1}{3}} dx\\ f \amp = \ln x \implies f^\prime = \frac{1}{x}\\ g^\prime \amp = x^\frac{-1}{3} \implies g = \frac{3x^{\frac{2}{3}}}{2}\\ \amp = \left. \frac{3x^{\frac{2}{3}}}{2} \ln x \right|_1^\infty - \int_1^\infty \frac{3x^{\frac{2}{3}}}{2} \frac{1}{x} dx \\ \amp = \lim_{a \rightarrow \infty} \left[ \frac{3a^{\frac{2}{3}} \ln a}{2} - \frac{3}{2} 1^{\frac{3}{2}} \ln 1 - \frac{3}{2} \int_1^a x^{-\frac{1}{3}} dx \right]\\ \amp = \lim_{a \rightarrow \infty} \frac{3}{2} \left[ a^{\frac{2}{3}} \ln a - \left. \frac{3x^{\frac{2}{3}}}{2} \right|_1^a \right] = \lim_{a \rightarrow \infty} \frac{3}{2}\left[ a^{\frac{2}{3}} \ln a - \frac{3a^{\frac{2}{3}}}{2} + \frac{3}{2} \right]\\ \amp = \lim_{a \rightarrow \infty} \frac{3}{2} \left[ a^{\frac{2}{3}} \left( \ln a - \frac{3}{2} \right) + \frac{3}{2} \right] = \infty \end{align*}

The integral diverges, so the series must as well.

\begin{equation*} \sum_{n=2}^\infty \frac{1}{n \ln n} \end{equation*}

We use the integral text again. In the integral, we use the substitution \(u = \ln x\text{.}\)

\begin{equation*} \int_2^\infty \frac{1}{x \ln x} dx = \int_{\ln 2}^\infty \frac{1}{u} du = \ln u \Big|_{\ln 2}^\infty = \infty \end{equation*}

The integral is divergent, so the sum is divergent as well. Also, note the following inequality.

\begin{equation*} \frac{1}{n} > \frac{1}{n \ln n} > \frac{1}{n^p} \end{equation*}

It seems that comparison should be helpful with a series of this type. However, the inequality shows that this series is asymptotically between the harmonic series and the other convergent \(p\) series. In comparison, it's slightly larger than a convergent series and slightly smaller than a divergent series, which is entirely unhelpful.

\begin{equation*} \sum_{n=1}^\infty ne^{-n^2} \end{equation*}

We use the integral test again. In the integral, we use the substitution \(u =x^2\text{.}\)

\begin{equation*} \int_1^\infty x e^{-x^2} dx = \frac{1}{2} \int_1^\infty e^{-u} du = \left. \frac{-1}{2} e^{-u} \right|_1^\infty = \frac{e}{2} \lt \infty \end{equation*}

The integral converges, so the sum does as well. Note that the sum does not have the value \(\frac{e}{2}\text{.}\)

\begin{equation*} \sum_{n=1}^\infty (-1)^n 3^{\frac{1}{n}} \end{equation*}

The root test is good for exponents.

\begin{equation*} \lim_{n \rightarrow \infty} \sqrt[n]{3^{\frac{1}{n}}} = \lim_{n \rightarrow \infty} 3^{\frac{1}{2n}} = 1 \end{equation*}

This limit is 1, so the test is inconclusive. Instead of using the root test, look at the limit of the terms. That limit is \(\pm 1\text{,}\) which is not zero, so the series must diverge by the test for divergence.

\begin{equation*} \sum_{n=1}^\infty \left( \frac{n}{n+1} \right)^{n^2} \end{equation*}

We use the root test again.

\begin{align*} \lim_{n \rightarrow \infty} \sqrt[n]{|a_n|} \amp = \lim_{n \rightarrow \infty} \sqrt[n]{\left( \frac{n}{n+1}\right)^{n^2} } = \lim_{n \rightarrow \infty} \left( \frac{n}{n+1} \right)^n = \frac{1}{\lim_{n \rightarrow \infty} \left( \frac{n+1}{n} \right)^n}\\ \amp = \frac{1}{\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n} = \frac{1}{e} \lt 1 \end{align*}

The limit is less than 1, so the series converges.

\begin{equation*} \sum_{n=1}^\infty \tan \left( \frac{1}{n} \right) \end{equation*}

There is an interesting comparison argument which we can use to tackle this difficult example. The derivative of tangent is \(\sec^2 x\text{.}\) Since \(\sec x\) is always \(>1\) or \(\lt -1\text{,}\) we have \(\sec^2x > 1\text{.}\) That is, the slope of tangent is always larger than 1. Since \(\tan 0 = 0\text{,}\) that means that near the origin, \(\tan x > x\text{.}\) Equivalently, for large \(n\text{,}\) \(\tan \frac{1}{n} > \frac{1}{n}\text{.}\) This allows us to compare our series to the harmonic series: our terms are larger than the harmonic series and the harmonic series diverges, so this series also diverges.